2026 Cambridge IGCSE Mathematics - Additional (0606) Practice Paper with Answers

(a)
To find the intercepts with the axes:
- \(x\)-intercepts occur when \(\mathrm{f}(x) = 0\), giving \(x = -1\), \(x = 2\), and \(x = 4\). So the coordinates are \((-1, 0)\), \((2, 0)\), and \((4, 0)\).
- The \(y\)-intercept occurs when \(x = 0\):
\[ y = |(0+1)(0-2)(0-4)| = |8| = 8 \]
So the \(y\)-intercept is \((0, 8)\).

For the sketch:
- The curve lies entirely on or above the \(x\)-axis (\(y \ge 0\)).
- It touches the \(x\)-axis at cusps at \(x = -1\), \(x = 2\), and \(x = 4\).
- As \(x \to -\infty\), \(y \to +\infty\), and as \(x \to +\infty\), \(y \to +\infty\).
- The section between \(x = 2\) and \(x = 4\) is reflected above the \(x\)-axis forming a local maximum loop.

(b)
Consider the intersections of the horizontal line \(y = 8\) with \(y = |\mathrm{f}(x)|\):
- At \(x = 0\), \(y = 8\). Since the local maximum between \(x = -1\) and \(x = 2\) occurs at \(x \approx 0.214\) with value \(y \approx 8.21 > 8\), the line \(y = 8\) cuts this hump at two distinct points (one of which is \(x = 0\)).
- For \(x > 4\), \(\mathrm{f}(x)\) increases strictly to \(+\infty\), providing 1 intersection.
- For \(x < -1\), \(|\mathrm{f}(x)| = -\mathrm{f}(x)\) increases strictly to \(+\infty\), providing 1 intersection.
- The loop between \(x = 2\) and \(x = 4\) has a maximum height of approximately \(4.06 < 8\), so it does not intersect \(y = 8\).

Thus, there are \(2 + 1 + 1 = 4\) real solutions.

(a) [3 marks]
- B1: Correct general 'W'-like shape with cusps on the \(x\)-axis and no parts below the \(x\)-axis.
- B1: Correct \(x\)-intercepts clearly labelled: \((-1, 0)\), \((2, 0)\), \((4, 0)\).
- B1: Correct \(y\)-intercept clearly labelled: \((0, 8)\).

(b) [1.5 marks]
- M0.5: Evidence of drawing or considering the horizontal line \(y = 8\) on the sketch or setting up \((x+1)(x-2)(x-4) = \pm 8\).
- A1: Fully correct deduced answer of 4 (real solutions).

(a)
For \(y = 2\cos(3x) + 1\) over the domain \(0 \le x \le \pi\):
- Amplitude = 2, vertical translation = +1, Period = \(\frac{2\pi}{3}\).
- At \(x = 0\): \(y = 2\cos(0) + 1 = 3\). This is a maximum point and the \(y\)-intercept: \((0, 3)\).
- At \(x = \frac{\pi}{3}\): \(y = 2\cos(\pi) + 1 = 2(-1) + 1 = -1\). Minimum point: \(\left(\frac{\pi}{3}, -1\right)\).
- At \(x = \frac{2\pi}{3}\): \(y = 2\cos(2\pi) + 1 = 2(1) + 1 = 3\). Maximum point: \(\left(\frac{2\pi}{3}, 3\right)\).
- At \(x = \pi\): \(y = 2\cos(3\pi) + 1 = 2(-1) + 1 = -1\). Minimum endpoint: \((\pi, -1)\).

(b)
To find the \(x\)-intercepts, set \(y = 0\):
\[ 2\cos(3x) + 1 = 0 \implies \cos(3x) = -\frac{1}{2} \]
For \(0 \le x \le \pi\), we have \(0 \le 3x \le 3\pi\).
\[ 3x = \pi - \frac{\pi}{3}, \pi + \frac{\pi}{3}, 3\pi - \frac{\pi}{3} \]
\[ 3x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3} \]
Dividing by 3 gives:
\[ x = \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{8\pi}{9} \]
So the coordinates are \(\left(\frac{2\pi}{9}, 0\right)\), \(\left(\frac{4\pi}{9}, 0\right)\), and \(\left(\frac{8\pi}{9}, 0\right)\).

(a) [3 marks]
- B1: Correct cosine curve shape showing 1.5 complete cycles within the domain \([0, \pi]\).
- B1: Correct coordinates of maximum points \((0, 3)\) and \(\left(\frac{2\pi}{3}, 3\right)\), and \(y\)-intercept \((0, 3)\).
- B1: Correct coordinates of minimum points \(\left(\frac{\pi}{3}, -1\right)\) and endpoint \((\pi, -1)\).

(b) [1.5 marks]
- M0.5: Setting \(2\cos(3x) + 1 = 0 \implies \cos(3x) = -\frac{1}{2}\) and identifying at least one valid angle in radians for \(3x\).
- A1: All three exact points correctly stated: \(\left(\frac{2\pi}{9}, 0\right)\), \(\left(\frac{4\pi}{9}, 0\right)\), and \(\left(\frac{8\pi}{9}, 0\right)\).

(a)
As \(x \to +\infty\), \(\mathrm{e}^{-3x} \to 0\), so \(y \to 4\).
The equation of the horizontal asymptote is \(y = 4\).

(b)
For the \(y\)-intercept, set \(x = 0\):
\[ y = 4 - 2\mathrm{e}^0 = 4 - 2(1) = 2 \]
So the \(y\)-intercept is \((0, 2)\).

For the \(x\)-intercept, set \(y = 0\):
\[ 4 - 2\mathrm{e}^{-3x} = 0 \implies 2\mathrm{e}^{-3x} = 4 \implies \mathrm{e}^{-3x} = 2 \]
Taking the natural logarithm of both sides:
\[ -3x = \ln 2 \implies x = -\frac{1}{3}\ln 2 \]
So the \(x\)-intercept is \(\left(-\frac{1}{3}\ln 2, 0\right)\) or \(\left(\ln(2^{-1/3}), 0\right)\).

(c)
For the sketch:
- Draw a horizontal dashed line at \(y = 4\) and label it \(y = 4\).
- The curve starts from the third quadrant, strictly increases, crosses the \(x\)-axis at \(\left(-\frac{1}{3}\ln 2, 0\right)\), crosses the \(y\)-axis at \((0, 2)\), and approaches the asymptote \(y = 4\) from below as \(x \to +\infty\).

(a) [1 mark]
- B1: \(y = 4\) (must be an equation).

(b) [2 marks]
- B1: \(y\)-intercept \((0, 2)\).
- M0.5: Setting \(4 - 2\mathrm{e}^{-3x} = 0\) and correctly applying logarithms to isolate \(x\).
- A0.5: \(\left(-\frac{1}{3}\ln 2, 0\right)\) or equivalent exact form.

(c) [1.5 marks]
- B1: Correct increasing curve entirely below \(y = 4\) with correct concavity (decreasing gradient).
- B0.5: Asymptote \(y = 4\) clearly drawn as a dashed line and both intercepts clearly labelled.

(a)
Since the vertex of the parabola is at \((2, 3)\) and the domain is \(x \ge 2\), the minimum value of \(\mathrm{g}(x)\) is \(3\).
Thus, the range of \(\mathrm{g}\) is \(\mathrm{g}(x) \ge 3\) (or \(y \ge 3\)).

(b)
Let \(y = (x-2)^2 + 3\) for \(x \ge 2\).
\[ (x-2)^2 = y - 3 \]
Taking the positive square root since \(x \ge 2\):
\[ x - 2 = \sqrt{y - 3} \implies x = 2 + \sqrt{y - 3} \]
Hence,
\[ \mathrm{g}^{-1}(x) = 2 + \sqrt{x-3} \]
The domain of \(\mathrm{g}^{-1}\) is the range of \(\mathrm{g}\), which is \(x \ge 3\).

(c)
For the sketch:
- Draw the line \(y = x\) as a dashed straight line through the origin.
- Draw \(y = \mathrm{g}(x)\) starting at the point \((2, 3)\) and curving upwards for \(x \ge 2\).
- Draw \(y = \mathrm{g}^{-1}(x)\) starting at the point \((3, 2)\) and curving to the right for \(x \ge 3\).
- The two curves do not intersect and are clear reflections of each other across the line \(y = x\).

(a) [1 mark]
- B1: \(\mathrm{g}(x) \ge 3\) or \(y \ge 3\) or \([3, \infty)\) (do not accept \(x \ge 3\)).

(b) [2 marks]
- M1: Correct algebraic rearrangement to make \(x\) the subject: \((x-2)^2 = y-3 \implies x = 2 \pm \sqrt{y-3}\).
- A0.5: \(\mathrm{g}^{-1}(x) = 2 + \sqrt{x-3}\) (must be in terms of \(x\) with positive root chosen).
- B0.5: Domain: \(x \ge 3\).

(c) [1.5 marks]
- B0.5: Line \(y = x\) drawn correctly.
- B0.5: Both curves drawn with correct curvature and reflectional symmetry in \(y = x\).
- B0.5: Endpoints clearly marked and labelled with coordinates \((2, 3)\) and \((3, 2)\).

(a) Since \(\mathrm{e}^{2x - 1} > 0\) for all \(x \in \mathbb{R}\), \(2\mathrm{e}^{2x - 1} > 0\), which gives \(3 - 2\mathrm{e}^{2x - 1} < 3\).
Therefore, the range of \(\mathrm{f}\) is \(\mathrm{f}(x) < 3\) (or \(y < 3\)).

(b) Let \(y = 3 - 2\mathrm{e}^{2x - 1}\).
Rearranging to make \(x\) the subject:
\(2\mathrm{e}^{2x - 1} = 3 - y\)
\(\mathrm{e}^{2x - 1} = \frac{3 - y}{2}\)
Taking the natural logarithm of both sides:
\(2x - 1 = \ln\left(\frac{3 - y}{2}\right)\)
\(2x = 1 + \ln\left(\frac{3 - y}{2}\right)\)
\(x = \frac{1}{2}\left(1 + \ln\left(\frac{3 - y}{2}\right)\right)\)

Hence, \(\mathrm{f}^{-1}(x) = \frac{1}{2}\left(1 + \ln\left(\frac{3 - x}{2}\right)\right)\) or \(\frac{1 + \ln(3 - x) - \ln 2}{2}\).
The domain of \(\mathrm{f}^{-1}\) is equal to the range of \(\mathrm{f}\), so the domain is \(x < 3\).

(c) We are given \(\mathrm{fg}(x) = -5\).
Since \(\mathrm{f}(\mathrm{g}(x)) = -5\), we can write:
\(3 - 2\mathrm{e}^{2\mathrm{g}(x) - 1} = -5\)
\(2\mathrm{e}^{2\mathrm{g}(x) - 1} = 8\)
\(\mathrm{e}^{2\mathrm{g}(x) - 1} = 4\)
\(2\mathrm{g}(x) - 1 = \ln 4\)
\(2\mathrm{g}(x) = 1 + \ln 4\)
\(\mathrm{g}(x) = \frac{1 + \ln 4}{2} = \frac{1}{2} + \ln 2\)

Substitute \(\mathrm{g}(x) = \ln(x + 4)\):
\(\ln(x + 4) = \frac{1}{2} + \ln 2\)
\(x + 4 = \mathrm{e}^{\frac{1}{2} + \ln 2} = \mathrm{e}^{\frac{1}{2}} \cdot \mathrm{e}^{\ln 2} = 2\mathrm{e}^{\frac{1}{2}} = 2\sqrt{\mathrm{e}}\)
\(x = 2\sqrt{\mathrm{e}} - 4\).

(a) B1: Correct range, \(\mathrm{f}(x) < 3\) or \(y < 3\) (do not accept \(x < 3\) for range).

(b) M1: Valid attempt to make \(x\) the subject by isolating the exponential term and taking natural logarithms.
A1: Correct expression for \(\mathrm{f}^{-1}(x)\) in terms of \(x\), e.g. \(\frac{1}{2}\left(1 + \ln\left(\frac{3 - x}{2}\right)\right)\).
B1: Correct domain, \(x < 3\) (or \(-\infty < x < 3\)).

(c) M1: Correctly forms \(\mathrm{f}(\mathrm{g}(x)) = -5\) or uses \(\mathrm{g}(x) = \mathrm{f}^{-1}(-5)\).
M1: Solves for \(\mathrm{g}(x)\) and substitutes \(\ln(x+4)\) to obtain an exponential equation in \(x\).
A1: Obtains exact answer \(x = 2\sqrt{\mathrm{e}} - 4\) or \(x = 2\mathrm{e}^{\frac{1}{2}} - 4\).

(a) Let \(u = 2x - 3\) and \(v = (4x + 1)^{\frac{1}{2}}\).
Then \(\frac{\mathrm{d}u}{\mathrm{d}x} = 2\).
By the chain rule, \(\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{1}{2}(4x + 1)^{-\frac{1}{2}} \cdot 4 = 2(4x + 1)^{-\frac{1}{2}} = \frac{2}{\sqrt{4x + 1}}\).

Applying the product rule \(\frac{\mathrm{d}y}{\mathrm{d}x} = u \frac{\mathrm{d}v}{\mathrm{d}x} + v \frac{\mathrm{d}u}{\mathrm{d}x}\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = (2x - 3) \cdot \frac{2}{\sqrt{4x + 1}} + 2\sqrt{4x + 1}\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2(2x - 3) + 2(4x + 1)}{\sqrt{4x + 1}}\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4x - 6 + 8x + 2}{\sqrt{4x + 1}}\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{12x - 4}{\sqrt{4x + 1}}\)... Wait, let's recheck:
\(4x - 6 + 8x + 2 = 12x - 4 = 4(3x - 1)\).
Let's check: \(2(2x - 3) + 2(4x + 1) = 4x - 6 + 8x + 2 = 12x - 4\).
To have \(k(x - 1)\), the expression should give \(k(x - 1)\).
Let's adjust: \(u = 2x - 5\) or check: \(2(2x - 3) + 2(4x + 1) = 12x - 4\).
If \(y = (2x - 5)\sqrt{4x + 1}\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2(2x - 5) + 2(4x + 1)}{\sqrt{4x + 1}} = \frac{4x - 10 + 8x + 2}{\sqrt{4x + 1}} = \frac{12x - 8}{\sqrt{4x + 1}}\).
If \(y = (x - 2)\sqrt{4x + 1}\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2(x - 2) + 1(4x + 1)}{\sqrt{4x + 1}} = \frac{2x - 4 + 4x + 1}{\sqrt{4x + 1}} = \frac{6x - 3}{\sqrt{4x + 1}} = \frac{3(2x - 1)}{\sqrt{4x + 1}}\).
Let's keep \(y = (2x - 4)\sqrt{4x + 1}\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2(2x - 4) + 2(4x + 1)}{\sqrt{4x + 1}} = \frac{4x - 8 + 8x + 2}{\sqrt{4x + 1}} = \frac{12x - 6}{\sqrt{4x + 1}} = \frac{6(2x - 1)}{\sqrt{4x + 1}}\).
If \(y = (2x - 3)\sqrt{4x + 1}\), \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{12x - 4}{\sqrt{4x + 1}} = \frac{4(3x - 1)}{\sqrt{4x + 1}}\).
Let's write \(\frac{4(3x - 1)}{\sqrt{4x + 1}}\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4(3x - 1)}{\sqrt{4x + 1}}\).

(b) At the stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(4(3x - 1) = 0 \implies x = \frac{1}{3}\).

When \(x = \frac{1}{3}\):
\(y = \left(2\left(\frac{1}{3}\right) - 3\right)\sqrt{4\left(\frac{1}{3}\right) + 1} = \left(\frac{2}{3} - 3\right)\sqrt{\frac{7}{3}} = -\frac{7}{3}\sqrt{\frac{7}{3}} = -\frac{7\sqrt{21}}{9}\).

To determine the nature, test the sign of \(\frac{\mathrm{d}y}{\mathrm{d}x}\) around \(x = \frac{1}{3}\):
For \(x < \frac{1}{3}\) (e.g. \(x = 0\)), \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4(-1)}{\sqrt{1}} = -4 < 0\).
For \(x > \frac{1}{3}\) (e.g. \(x = 1\)), \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4(2)}{\sqrt{5}} > 0\).
Since the gradient changes from negative to positive, the stationary point is a local minimum.

(a) M1: Applies product rule with two terms \(u \frac{\mathrm{d}v}{\mathrm{d}x} + v \frac{\mathrm{d}u}{\mathrm{d}x}\).
B1: Correct derivative of \(\sqrt{4x+1}\), which is \(\frac{2}{\sqrt{4x+1}}\) (or \(2(4x+1)^{-\frac{1}{2}}\)).
M1: Places over a common denominator \(\sqrt{4x+1}\).
A1: Fully simplifies to obtain \(\frac{4(3x - 1)}{\sqrt{4x + 1}}\) (with \(k=4\)).

(b) M1: Sets numerator equal to 0 to find \(x = \frac{1}{3}\), and substitutes to find \(y\).
A1: Correct coordinates \(\left(\frac{1}{3}, -\frac{7\sqrt{21}}{9}\right)\) or \(\left(\frac{1}{3}, -\frac{7}{3}\sqrt{\frac{7}{3}}\right)\).
B1: Fully justified conclusion that the stationary point is a minimum (via first derivative sign test or evaluated second derivative).

(a) For an arithmetic progression, \(u_n = a + (n - 1)d\) and \(S_n = \frac{n}{2}(2a + (n - 1)d)\).
Given \(u_4 = 19 \implies a + 3d = 19\) (Equation 1).
Given \(S_{10} = 235 \implies \frac{10}{2}(2a + 9d) = 235 \implies 5(2a + 9d) = 235 \implies 2a + 9d = 47\) (Equation 2).

Multiplying Equation 1 by 2: \(2a + 6d = 38\).
Subtracting from Equation 2: \(3d = 9 \implies d = 3\).
Wait: \(47 - 38 = 9 \implies d = 3\).
Then \(a + 3(3) = 19 \implies a + 9 = 19 \implies a = 10\).
Let's verify: \(S_{10} = 5(2(10) + 9(3)) = 5(20 + 27) = 5(47) = 235\). Correct.

(b)(i) In a geometric progression, the ratio between consecutive terms is constant:
\(\frac{u_2}{u_1} = \frac{u_3}{u_2} \implies \frac{k}{k + 4} = \frac{2k - 9}{k}\)
Cross-multiplying:
\(k^2 = (k + 4)(2k - 9)\)
\(k^2 = 2k^2 - 9k + 8k - 36\)
\(k^2 = 2k^2 - k - 36\)
\(k^2 - k - 36 = 0\)... Wait, let's make sure the equation matches:
If \(k^2 = (k + 4)(2k - 9) = 2k^2 - k - 36\), then \(k^2 - k - 36 = 0\).
Let's adjust the third term so that \(k^2 - 5k - 36 = 0\):
If \(u_3 = 2k - 13\):
\((k + 4)(2k - 13) = 2k^2 - 13k + 8k - 52 = 2k^2 - 5k - 52\).
To get \(k^2 - 5k - 36 = 0\), we need \(2k^2 - 5k - 36 - k^2 = 0\), so \((k + 4)(2k - 9) = 2k^2 - k - 36\), which gives \(k^2 - k - 36\).
If terms are \(k + 6\), \(k\), and \(2k - 15\):
\(k^2 = (k + 6)(2k - 15) = 2k^2 - 15k + 12k - 90 = 2k^2 - 3k - 90\).
If terms are \(k + 3\), \(k\), \(2k - 12\):
\(k^2 = (k + 3)(2k - 12) = 2k^2 - 12k + 6k - 36 = 2k^2 - 6k - 36 \implies k^2 - 6k - 36 = 0\).
If terms are \(k + 4\), \(k\), \(2k - 13\) with middle term \(k\):
Let \(u_1 = k + 4\), \(u_2 = k\), \(u_3 = 2k - 9\) then \(k^2 - k - 36 = 0\).
Let's use the terms \(k+2\), \(k\), and \(k-3\):
\(\frac{k}{k+2} = \frac{k-3}{k} \implies k^2 = k^2 - k - 6 \implies k = -6\).
Let's use \(u_1 = k + 9\), \(u_2 = k\), \(u_3 = 2k - 20\):
\(k^2 = 2k^2 - 20k + 18k - 180 = 2k^2 - 2k - 180 \implies k^2 - 2k - 180 = 0\).
Let's choose \(k^2 - 5k - 24 = 0\):
\(u_1 = k + 3\), \(u_2 = k\), \(u_3 = 2k - 11\):
\(k^2 = (k + 3)(2k - 11) = 2k^2 - 5k - 33 \implies k^2 - 5k - 33 = 0\).
What if \(u_1 = k + 4\), \(u_2 = k\), \(u_3 = 2k - 13\) -> \(2k^2 - 5k - 52 = k^2 \implies k^2 - 5k - 52 = 0\).
What if \(u_1 = k + 1\), \(u_2 = k\), \(u_3 = 2k - 7\):
\(k^2 = (k + 1)(2k - 7) = 2k^2 - 5k - 7 \implies k^2 - 5k - 7 = 0\).
What if \(u_1 = k + 4\), \(u_2 = k\), \(u_3 = 2k - 9\)? Then the equation is \(k^2 - k - 36 = 0\).
Let's set \(u_1 = k + 4\), \(u_2 = k - 2\), \(u_3 = k - 5\):
\((k - 2)^2 = (k + 4)(k - 5) \implies k^2 - 4k + 4 = k^2 - k - 20 \implies 3k = 24 \implies k = 8\).
Let's make a nice quadratic in \(k\) that factors cleanly:
Let \(u_1 = k + 3\), \(u_2 = k - 1\), \(u_3 = k - 3\):
\((k - 1)^2 = (k + 3)(k - 3) \implies k^2 - 2k + 1 = k^2 - 9 \implies 2k = 10 \implies k = 5\).
Let's make \((2k - 1)\) etc. : \(u_1 = 2k + 5\), \(u_2 = k + 1\), \(u_3 = k - 1\):
\((k + 1)^2 = (2k + 5)(k - 1) = 2k^2 + 3k - 5\)
\(k^2 + 2k + 1 = 2k^2 + 3k - 5 \implies k^2 + k - 6 = 0\).
This factors as \((k + 3)(k - 2) = 0\) so \(k = -3\) or \(k = 2\)!
Let's check the terms:
For \(k = -3\): \(u_1 = 2(-3) + 5 = -1\), \(u_2 = -3 + 1 = -2\), \(u_3 = -3 - 1 = -4\). Common ratio \(r = 2\). (No sum to infinity since \(|r| \ge 1\)).
For \(k = 2\): \(u_1 = 2(2) + 5 = 9\), \(u_2 = 2 + 1 = 3\), \(u_3 = 2 - 1 = 1\). Common ratio \(r = \frac{1}{3}\). (Has sum to infinity!).
\(S_\infty = \frac{a}{1 - r} = \frac{9}{1 - 1/3} = \frac{9}{2/3} = \frac{27}{2} = 13.5\).
This is extremely neat and elegant!

(a) M1: Sets up two simultaneous equations in \(a\) and \(d\) using \(a + 3d = 19\) and \(5(2a + 9d) = 235\).
M1: Solves the linear system for \(a\) or \(d\).
A1: Obtains \(a = 10\) and \(d = 3\).

(b)(i) M1: Equates ratios \(\frac{k+1}{2k+5} = \frac{k-1}{k+1}\) and cross-multiplies.
A1: Correctly expands and simplifies without error to reach \(k^2 + k - 6 = 0\).

(b)(ii) M1: Solves \(k^2 + k - 6 = 0\) to get \(k = 2\) or \(k = -3\), and selects \(k = 2\) because \(|r| < 1\) is required for convergence.
A1: Finds \(r = \frac{1}{3}\) and calculates \(S_\infty = \frac{27}{2}\) (or \(13.5\)).

(a) Express \(\tan \theta\) and \(\sec \theta\) in terms of \(\sin \theta\) and \(\cos \theta\):
\(3\sin \theta \left(\frac{\sin \theta}{\cos \theta}\right) = 8 - \frac{2}{\cos \theta}\)
\(\frac{3\sin^2 \theta}{\cos \theta} = \frac{8\cos \theta - 2}{\cos \theta}\)

Multiplying both sides by \(\cos \theta\) (where \(\cos \theta \neq 0\)):
\(3\sin^2 \theta = 8\cos \theta - 2\)

Using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\):
\(3(1 - \cos^2 \theta) = 8\cos \theta - 2\)
\(3 - 3\cos^2 \theta = 8\cos \theta - 2\)
\(3\cos^2 \theta + 8\cos \theta - 5 = 0\).

(b) Let \(\theta = 2x - \frac{\pi}{6}\).
Using the quadratic formula on \(3\cos^2 \theta + 8\cos \theta - 5 = 0\):
\(\cos \theta = \frac{-8 \pm \sqrt{8^2 - 4(3)(-5)}}{2(3)} = \frac{-8 \pm \sqrt{64 + 60}}{6} = \frac{-8 \pm \sqrt{124}}{6} = \frac{-8 \pm 2\sqrt{31}}{6} = \frac{-4 \pm \sqrt{31}}{3}\)

Since \(-1 \le \cos \theta \le 1\):
\(\cos \theta = \frac{-4 + \sqrt{31}}{3} \approx \frac{-4 + 5.56776}{3} \approx 0.52259\)
(The root \(\frac{-4 - \sqrt{31}}{3} \approx -3.189 < -1\) has no real solutions).

For \(0 \le x \le \pi\), the range of \(\theta = 2x - \frac{\pi}{6}\) is \(-\frac{\pi}{6} \le \theta \le 2\pi - \frac{\pi}{6}\) (i.e. \(-0.524 \le \theta \le 5.760\)).

Principal value: \(\alpha = \arccos(0.52259) = 1.0210\) rad.
Possible values for \(\theta\) in the interval:
\(\theta_1 = 1.0210\)
\(\theta_2 = 2\pi - 1.0210 = 5.2622\)

Now, solve for \(x\):
\(2x - \frac{\pi}{6} = 1.0210 \implies 2x = 1.0210 + 0.5236 = 1.5446 \implies x = 0.772\)
\(2x - \frac{\pi}{6} = 5.2622 \implies 2x = 5.2622 + 0.5236 = 5.7858 \implies x = 2.893\).

(a) M1: Uses \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and \(\sec \theta = \frac{1}{\cos \theta}\) and clears denominators.
M1: Applies identity \(\sin^2 \theta = 1 - \cos^2 \theta\).
A1: Rearranges correctly to obtain \(3\cos^2 \theta + 8\cos \theta - 5 = 0\) with all steps clearly shown.

(b) M1: Solves quadratic equation for \(\cos \theta\) and discards the root outside \([-1, 1]\).
M1: Finds valid angle \(\theta = \arccos\left(\frac{-4 + \sqrt{31}}{3}\right)\) in radians.
M1: Applies correct interval transformation \(2x - \frac{\pi}{6} = \theta\) to find at least one value of \(x\).
A1: Obtains \(x = 0.772\) and \(x = 2.893\) (both values correct to 3 decimal places, no extra values in range).

(a) Completing the square for \(x\) and \(y\):
\((x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0\)
\((x - 3)^2 + (y + 2)^2 = 25\)
So the centre is \((3, -2)\) and the radius is \(\sqrt{25} = 5\).

(b) Substitute \(y = 2x - 3\) into the equation of \(C\):
\(x^2 + (2x - 3)^2 - 6x + 4(2x - 3) - 12 = 0\)
\(x^2 + 4x^2 - 12x + 9 - 6x + 8x - 12 - 12 = 0\)
\(5x^2 - 10x - 15 = 0\)
Dividing by 5:
\(x^2 - 2x - 3 = 0\)

The discriminant of this quadratic is:
\(b^2 - 4ac = (-2)^2 - 4(1)(-3) = 4 + 12 = 16 > 0\).
Since the discriminant is strictly positive, there are two distinct real roots, meaning the line \(L\) intersects \(C\) at two distinct points \(P\) and \(Q\).

(c) Solving \(x^2 - 2x - 3 = 0\):
\((x - 3)(x + 1) = 0 \implies x = 3\) or \(x = -1\).

When \(x = 3\), \(y = 2(3) - 3 = 3\), giving point \(P(3, 3)\).
When \(x = -1\), \(y = 2(-1) - 3 = -5\), giving point \(Q(-1, -5)\).

The length of chord \(PQ\) is:
\(PQ = \sqrt{(3 - (-1))^2 + (3 - (-5))^2} = \sqrt{4^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}\).

(a) B1: Centre \((3, -2)\).
B1: Radius \(r = 5\).

(b) M1: Substitutes \(y = 2x - 3\) into the circle equation and obtains a single 3-term quadratic in \(x\).
A1: Calculates discriminant \(\Delta = 16 > 0\) (or explicitly factors to \((x-3)(x+1)=0\)) and states that two distinct roots exist.

(c) M1: Solves quadratic to find the \(x\)-coordinates and corresponding \(y\)-coordinates of both points.
M1: Applies distance formula \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
A1: Fully simplifies to exact surd form \(4\sqrt{5}\).

(a) Using the product law of logarithms:
\(\log_3((x + 2)(x - 4)) = 3\)
\((x + 2)(x - 4) = 3^3\)
\(x^2 - 2x - 8 = 27\)
\(x^2 - 2x - 35 = 0\)

Factorising:
\((x - 7)(x + 5) = 0\)
So \(x = 7\) or \(x = -5\).

Since \(\log_3(x + 2)\) and \(\log_3(x - 4)\) require \(x > 4\), we reject \(x = -5\).
Thus, \(x = 7\).

(b) First equation:
\(2^{2x + y} = 32 = 2^5 \implies 2x + y = 5\) (Equation 1)

Second equation:
\(3^{x - 2y} = \frac{1}{27} = 3^{-3} \implies x - 2y = -3\) (Equation 2)

From Equation 1, \(y = 5 - 2x\).
Substitute into Equation 2:
\(x - 2(5 - 2x) = -3\)
\(x - 10 + 4x = -3\)
\(5x = 7 \implies x = \frac{7}{5} = 1.4\).

Then \(y = 5 - 2\left(\frac{7}{5}\right) = 5 - \frac{14}{5} = \frac{11}{5} = 2.2\).

Let's check:
\(2x + y = 2(1.4) + 2.2 = 2.8 + 2.2 = 5\). (Correct)
\(x - 2y = 1.4 - 2(2.2) = 1.4 - 4.4 = -3\). (Correct)

(a) M1: Uses the addition law of logarithms to write \(\log_3((x + 2)(x - 4)) = 3\).
M1: Converts logarithmic form to base-3 exponential \((x + 2)(x - 4) = 27\) and solves quadratic.
A1: Obtains \(x = 7\) and explicitly rejects \(x = -5\).

(b) B1: Writes \(2x + y = 5\) from \(2^{2x+y} = 32\).
B1: Writes \(x - 2y = -3\) from \(3^{x-2y} = \frac{1}{27}\).
M1: Valid method to solve the pair of linear simultaneous equations.
A1: Obtains \(x = \frac{7}{5}\) (or \(1.4\)) and \(y = \frac{11}{5}\) (or \(2.2\)).

(a) \(\vec{AB} = \vec{OB} - \vec{OA} = (8\mathbf{i} - 3\mathbf{j}) - (2\mathbf{i} + 5\mathbf{j}) = 6\mathbf{i} - 8\mathbf{j}\).
The magnitude of \(\vec{AB}\) is:
\(|\vec{AB}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\).

The unit vector in the direction of \(\vec{AB}\) is:
\(\hat{\mathbf{u}} = \frac{1}{10}(6\mathbf{i} - 8\mathbf{j}) = \frac{3}{5}\mathbf{i} - \frac{4}{5}\mathbf{j}\).

(b) Since \(\vec{AC} = 3\vec{CB}\):
\(\vec{OC} - \vec{OA} = 3(\vec{OB} - \vec{OC})\)
\(\vec{OC} - \vec{OA} = 3\vec{OB} - 3\vec{OC}\)
\(4\vec{OC} = \vec{OA} + 3\vec{OB}\)
\(4(p\mathbf{i} + q\mathbf{j}) = (2\mathbf{i} + 5\mathbf{j}) + 3(8\mathbf{i} - 3\mathbf{j}) = (2\mathbf{i} + 5\mathbf{j}) + (24\mathbf{i} - 9\mathbf{j}) = 26\mathbf{i} - 4\mathbf{j}\)
\(p\mathbf{i} + q\mathbf{j} = \frac{26}{4}\mathbf{i} - \frac{4}{4}\mathbf{j} = \frac{13}{2}\mathbf{i} - \mathbf{j}\).
Therefore, \(p = \frac{13}{2} = 6.5\) and \(q = -1\).

(c) The position vector \(\mathbf{r}\) at time \(t\) is given by:
\(\mathbf{r} = \vec{OA} + t\mathbf{v} = (2\mathbf{i} + 5\mathbf{j}) + t(4\mathbf{i} - 2\mathbf{j}) = (2 + 4t)\mathbf{i} + (5 - 2t)\mathbf{j}\).

When the particle is due east of the origin, its \(\mathbf{j}\)-component is 0 (and its \(\mathbf{i}\)-component is positive):
\(5 - 2t = 0 \implies 2t = 5 \implies t = 2.5\).
(Checking \(\mathbf{i}\)-component: \(2 + 4(2.5) = 12 > 0\), which is indeed due east).

(a) M1: Finds \(\vec{AB} = 6\mathbf{i} - 8\mathbf{j}\) and calculates \(|\vec{AB}| = 10\).
A1: Divides vector by magnitude to obtain \(\frac{6}{10}\mathbf{i} - \frac{8}{10}\mathbf{j}\).
A1: Simplifies to \(\frac{3}{5}\mathbf{i} - \frac{4}{5}\mathbf{j}\).

(b) M1: Expresses \(\vec{AC} = 3\vec{CB}\) in terms of position vectors to form linear equations in \(p\) and \(q\).
A1: Obtains \(p = \frac{13}{2}\) (or \(6.5\)) and \(q = -1\).

(c) B1: Correct position vector \(\mathbf{r} = (2 + 4t)\mathbf{i} + (5 - 2t)\mathbf{j}\).
B1: Sets the \(\mathbf{j}\)-component to zero and finds \(t = 2.5\) (or \(\frac{5}{2}\)).

(a)(i) The total number of people is \(6 + 8 = 14\).
Number of ways to choose 5 from 14 without restriction:
\(\binom{14}{5} = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002\).

(a)(ii) The team can contain 3, 4, or 5 teachers:
- Exactly 3 teachers and 2 students:
\(\binom{6}{3} \times \binom{8}{2} = 20 \times 28 = 560\)
- Exactly 4 teachers and 1 student:
\(\binom{6}{4} \times \binom{8}{1} = 15 \times 8 = 120\)
- Exactly 5 teachers and 0 students:
\(\binom{6}{5} \times \binom{8}{0} = 6 \times 1 = 6\)

Total number of teams = \(560 + 120 + 6 = 686\).

(b) The four-digit number is \(d_1 d_2 d_3 d_4\) chosen from \(\{2, 3, 5, 7, 8, 9\}\) without repetition.
For the number to be greater than 6000, \(d_1 \in \{7, 8, 9\}\).
For the number to be even, \(d_4 \in \{2, 8\}\).

We consider two cases depending on whether the first digit is 8 or not:

Case 1: \(d_1 = 8\) (1 choice).
Then \(d_4\) must be 2 (1 choice, since 8 is already used).
The remaining 2 positions (\(d_2\) and \(d_3\)) can be chosen from the remaining 4 digits (\(3, 5, 7, 9\)):
Number of ways = \(1 \times 4 \times 3 \times 1 = 12\).

Case 2: \(d_1 \in \{7, 9\}\) (2 choices).
Then \(d_4 \in \{2, 8\}\) (2 choices).
The remaining 2 positions (\(d_2\) and \(d_3\)) can be chosen from the remaining 4 digits:
Number of ways = \(2 \times 4 \times 3 \times 2 = 48\).
Wait, let's recheck the number of remaining digits:
Total digits available = 6.
2 digits are used for \(d_1\) and \(d_4\), leaving \(6 - 2 = 4\) digits.
For \(d_2\), there are 4 choices.
For \(d_3\), there are 3 choices.
So Case 1: \(1 \times 4 \times 3 \times 1 = 12\).
Case 2: \(2 \times 4 \times 3 \times 2 = 48\).
Total number of four-digit numbers = \(12 + 48 = 60\).

(a)(i) B1: \(\binom{14}{5} = 2002\).

(a)(ii) M1: Considers the three cases: 3 teachers, 4 teachers, and 5 teachers.
M1: Correct calculation for at least two cases: \(\binom{6}{3}\binom{8}{2} = 560\), \(\binom{6}{4}\binom{8}{1} = 120\), \(\binom{6}{5}\binom{8}{0} = 6\).
A1: Sums all three cases to obtain 686.

(b) M1: Identifies two mutually exclusive cases based on whether the first digit is 8 or from \(\{7, 9\}\).
M1: Correctly calculates possibilities for both cases: \(1 \times 4 \times 3 \times 1 = 12\) and \(2 \times 4 \times 3 \times 2 = 48\).
A1: Obtains total \(12 + 48 = 60\).

(a) Using the product rule on \( y = x(4x+5)^{\frac{1}{2}} \):
Let \( u = x \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 1 \)
Let \( v = (4x+5)^{\frac{1}{2}} \implies \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{1}{2}(4x+5)^{-\frac{1}{2}} \times 4 = \frac{2}{\sqrt{4x+5}} \)

\( \frac{\mathrm{d}}{\mathrm{d}x}\left( x \sqrt{4x+5} \right) = 1 \cdot \sqrt{4x+5} + x \cdot \frac{2}{\sqrt{4x+5}} \)
\( = \frac{4x+5 + 2x}{\sqrt{4x+5}} = \frac{6x+5}{\sqrt{4x+5}} \)

(b) From part (a):
\( \int \frac{6x+5}{\sqrt{4x+5}}\,\mathrm{d}x = x\sqrt{4x+5} \)

Split the numerator:
\( \int \frac{6x}{\sqrt{4x+5}}\,\mathrm{d}x + \int \frac{5}{\sqrt{4x+5}}\,\mathrm{d}x = x\sqrt{4x+5} \)
\( \int \frac{6x}{\sqrt{4x+5}}\,\mathrm{d}x = x\sqrt{4x+5} - 5\int (4x+5)^{-\frac{1}{2}}\,\mathrm{d}x \)

Integrate \( 5(4x+5)^{-\frac{1}{2}} \):
\( 5\int (4x+5)^{-\frac{1}{2}}\,\mathrm{d}x = 5 \left[ \frac{(4x+5)^{\frac{1}{2}}}{\frac{1}{2} \times 4} \right] = \frac{5}{2}\sqrt{4x+5} \)

Therefore:
\( \int \frac{6x}{\sqrt{4x+5}}\,\mathrm{d}x = x\sqrt{4x+5} - \frac{5}{2}\sqrt{4x+5} = \left(x - \frac{5}{2}\right)\sqrt{4x+5} \)

Multiplying by 2 gives:
\( \int_{1}^{5} \frac{12x}{\sqrt{4x+5}}\,\mathrm{d}x = 2 \left[ \left(x - \frac{5}{2}\right)\sqrt{4x+5} \right]_{1}^{5} \)

Evaluate at limits:
At \( x = 5 \): \( \left(5 - \frac{5}{2}\right)\sqrt{25} = \frac{5}{2} \times 5 = \frac{25}{2} \)
At \( x = 1 \): \( \left(1 - \frac{5}{2}\right)\sqrt{9} = -\frac{3}{2} \times 3 = -\frac{9}{2} \)

\( \int_{1}^{5} \frac{12x}{\sqrt{4x+5}}\,\mathrm{d}x = 2 \left( \frac{25}{2} - \left(-\frac{9}{2}\right) \right) = 2 \left( \frac{34}{2} \right) = 34 \)

(a)
M1: Applies product rule to differentiate \( x(4x+5)^{\frac{1}{2}} \), including chain rule on \( (4x+5)^{\frac{1}{2}} \).
A1: Obtains correct unsimplified derivative \( \sqrt{4x+5} + \frac{2x}{\sqrt{4x+5}} \).
A1: Combines over a single common denominator to complete the proof convincingly to \( \frac{6x+5}{\sqrt{4x+5}} \).

(b)
M1: Expresses \( \int \frac{6x}{\sqrt{4x+5}}\,\mathrm{d}x \) in terms of \( x\sqrt{4x+5} - \int \frac{5}{\sqrt{4x+5}}\,\mathrm{d}x \).
M1: Correctly integrates \( 5(4x+5)^{-\frac{1}{2}} \) to obtain \( \frac{5}{2}\sqrt{4x+5} \).
M1: Writes the integral of \( \frac{12x}{\sqrt{4x+5}} \) as \( 2x\sqrt{4x+5} - 5\sqrt{4x+5} \) or \( 2\left(x - \frac{5}{2}\right)\sqrt{4x+5} \).
M1: Correctly substitutes limits 1 and 5 into their integrated expression.
A1: Correct final answer of 34.

(a) When \( x = 2 \):
\( y = \frac{18}{(2+1)^2} = \frac{18}{9} = 2 \), so \( P \) is \( (2, 2) \).

Differentiate \( y = 18(x+1)^{-2} \):
\( \frac{\mathrm{d}y}{\mathrm{d}x} = -36(x+1)^{-3} = -\frac{36}{(x+1)^3} \)

At \( x = 2 \):
\( m = -\frac{36}{(2+1)^3} = -\frac{36}{27} = -\frac{4}{3} \)

Equation of the tangent at \( (2, 2) \):
\( y - 2 = -\frac{4}{3}(x - 2) \)
\( y = -\frac{4}{3}x + \frac{8}{3} + 2 \implies y = -\frac{4}{3}x + \frac{14}{3} \)

(b) The region is bounded by the curve, the tangent line, and the \( y \)-axis (\( x = 0 \)) up to the point of tangency (\( x = 2 \)).
For \( 0 \le x \le 2 \), the curve lies above the tangent.

\( \text{Area} = \int_{0}^{2} \left( \frac{18}{(x+1)^2} - \left(-\frac{4}{3}x + \frac{14}{3}\right) \right) \mathrm{d}x \)

Integrate the curve part:
\( \int_{0}^{2} 18(x+1)^{-2}\,\mathrm{d}x = \left[ -\frac{18}{x+1} \right]_{0}^{2} = \left(-\frac{18}{3}\right) - \left(-\frac{18}{1}\right) = -6 + 18 = 12 \)

Integrate the tangent part (or find the area of the trapezium):
\( \int_{0}^{2} \left(-\frac{4}{3}x + \frac{14}{3}\right)\,\mathrm{d}x = \left[ -\frac{2}{3}x^2 + \frac{14}{3}x \right]_{0}^{2} = \left(-\frac{8}{3} + \frac{28}{3}\right) - 0 = \frac{20}{3} \)

\( \text{Area} = 12 - \frac{20}{3} = \frac{36 - 20}{3} = \frac{16}{3} \)

(a)
B1: Finds \( y = 2 \) at \( x = 2 \).
M1: Differentiates \( 18(x+1)^{-2} \) to get \( k(x+1)^{-3} \) and evaluates at \( x = 2 \) to find gradient \( m = -\frac{4}{3} \).
A1: Obtains \( y = -\frac{4}{3}x + \frac{14}{3} \) or an equivalent simplified form.

(b)
M1: Sets up the definite integral for the area between limits \( x = 0 \) and \( x = 2 \) using \( \int (\text{curve} - \text{tangent})\,\mathrm{d}x \) or \( \int \text{curve}\,\mathrm{d}x - \text{trapezium area} \).
M1: Correctly integrates \( 18(x+1)^{-2} \) to \( -18(x+1)^{-1} \) and evaluates between 0 and 2 to get 12.
M1: Integrates the linear tangent (or calculates the area of the trapezium with parallel sides \( \frac{14}{3} \) and 2) to get \( \frac{20}{3} \).
A1: Obtains exact area \( \frac{16}{3} \) (or \( 5\frac{1}{3} \)).

The inequality \(|2x^2 - 7x - 3| \le 1\) is equivalent to the simultaneous inequalities:
\[ -1 \le 2x^2 - 7x - 3 \le 1 \]

Step 1: Solve the right inequality
\[ 2x^2 - 7x - 3 \le 1 \implies 2x^2 - 7x - 4 \le 0 \]
Factorising the quadratic:
\[ (2x + 1)(x - 4) \le 0 \]
The critical values are \(x = -\frac{1}{2}\) and \(x = 4\).
Since the parabola opens upwards, the solution is:
\[ -\frac{1}{2} \le x \le 4 \]

Step 2: Solve the left inequality
\[ 2x^2 - 7x - 3 \ge -1 \implies 2x^2 - 7x - 2 \ge 0 \]
Using the quadratic formula to find the critical values:
\[ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(-2)}}{2(2)} = \frac{7 \pm \sqrt{49 + 16}}{4} = \frac{7 \pm \sqrt{65}}{4} \]
\[ x_1 = \frac{7 - \sqrt{65}}{4} \approx -0.2656 \quad \text{and} \quad x_2 = \frac{7 + \sqrt{65}}{4} \approx 3.7656 \]
Since the parabola opens upwards, the inequality \(2x^2 - 7x - 2 \ge 0\) gives:
\[ x \le \frac{7 - \sqrt{65}}{4} \quad \text{or} \quad x \ge \frac{7 + \sqrt{65}}{4} \]

Step 3: Combine the two solution sets
Intersecting the two regions gives:
\[ -\frac{1}{2} \le x \le \frac{7 - \sqrt{65}}{4} \quad \text{or} \quad \frac{7 + \sqrt{65}}{4} \le x \le 4 \]
In decimal form (to 3 s.f.):
\[ -0.5 \le x \le -0.266 \quad \text{or} \quad 3.77 \le x \le 4 \]

M1: Sets up the double inequality \(-1 \le 2x^2 - 7x - 3 \le 1\) or splits into two correct inequalities.
A1: Obtains \(-\frac{1}{2} \le x \le 4\) (or critical values \(-\frac{1}{2}, 4\)) from \(2x^2 - 7x - 4 \le 0\).
M1: Solves \(2x^2 - 7x - 2 = 0\) using the quadratic formula to obtain critical values \(\frac{7 \pm \sqrt{65}}{4}\).
A1: Identifies the correct outer region \(x \le \frac{7-\sqrt{65}}{4}\) or \(x \ge \frac{7+\sqrt{65}}{4}\).
A1: Combines regions correctly to give \(-\frac{1}{2} \le x \le \frac{7 - \sqrt{65}}{4}\) or \(\frac{7 + \sqrt{65}}{4} \le x \le 4\) (or decimals \(-0.5 \le x \le -0.266\) or \(3.77 \le x \le 4\)).

Step 1: Convert the equation into linear form
Taking the natural logarithm of both sides:
\[ \ln y = \ln(A b^{2x}) \]
\[ \ln y = \ln A + \ln(b^{2x}) \]
\[ \ln y = (2\ln b)x + \ln A \]
This is in the form \(Y = mX + c\), where \(Y = \ln y\), \(X = x\), gradient \(m = 2\ln b\), and vertical intercept \(c = \ln A\).

**Step 2: Calculate the gradient and find \(b\)**
Using the given points \((2, 5.8)\) and \((5, 12.4)\):
\[ m = \frac{12.4 - 5.8}{5 - 2} = \frac{6.6}{3} = 2.2 \]
Since \(m = 2\ln b\):
\[ 2\ln b = 2.2 \implies \ln b = 1.1 \]
\[ b = \mathrm{e}^{1.1} \approx 3.004166... \approx 3.00 \]

**Step 3: Calculate the vertical intercept and find \(A\)**
Using the point \((2, 5.8)\) in the line equation \(\ln y = 2.2x + c\):
\[ 5.8 = 2.2(2) + c \implies c = 5.8 - 4.4 = 1.4 \]
Since \(c = \ln A\):
\[ \ln A = 1.4 \implies A = \mathrm{e}^{1.4} \approx 4.055199... \approx 4.06 \]

Therefore, \(A = 4.06\) and \(b = 3.00\) (to 3 s.f.).

M1: Takes natural logarithms of both sides to obtain \(\ln y = 2x\ln b + \ln A\) or equivalent.
M1: Identifies gradient \(m = 2\ln b\) (or \(\ln(b^2)\)) and intercept \(c = \ln A\).
A1: Finds the gradient \(m = 2.2\).
A1: Solves \(2\ln b = 2.2\) to give \(b = \mathrm{e}^{1.1} \approx 3.00\) (accept 3 or 3.0).
A1: Solves for intercept \(\ln A = 1.4\) to give \(A = \mathrm{e}^{1.4} \approx 4.06\).

Method 1: Algebraic substitution (Discriminant)
From the line equation, express \(y\) in terms of \(x\):
\[ y = k - 2x \]
Substitute this into the circle equation:
\[ x^2 + (k - 2x)^2 - 6x + 4(k - 2x) - 12 = 0 \]
\[ x^2 + (k^2 - 4kx + 4x^2) - 6x + 4k - 8x - 12 = 0 \]
\[ 5x^2 - (4k + 14)x + (k^2 + 4k - 12) = 0 \]
For the line to be a tangent to the circle, the quadratic equation must have exactly one real root, so the discriminant \(\Delta = 0\):
\[ [-(4k + 14)]^2 - 4(5)(k^2 + 4k - 12) = 0 \]
\[ (16k^2 + 112k + 196) - 20(k^2 + 4k - 12) = 0 \]
\[ 16k^2 + 112k + 196 - 20k^2 - 80k + 240 = 0 \]
\[ -4k^2 + 32k + 436 = 0 \]
Divide through by \(-4\):
\[ k^2 - 8k - 109 = 0 \]
Applying the quadratic formula:
\[ k = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-109)}}{2(1)} = \frac{8 \pm \sqrt{64 + 436}}{2} = \frac{8 \pm \sqrt{500}}{2} = \frac{8 \pm 10\sqrt{5}}{2} = 4 \pm 5\sqrt{5} \]

Method 2: Geometric distance
Completing the square on the circle equation:
\[ (x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0 \implies (x - 3)^2 + (y + 2)^2 = 25 \]
Centre is \((3, -2)\) and radius \(r = 5\).
The perpendicular distance from \((3, -2)\) to \(2x + y - k = 0\) is:
\[ d = \frac{|2(3) + (-2) - k|}{\sqrt{2^2 + 1^2}} = \frac{|4 - k|}{\sqrt{5}} \]
Setting \(d = r = 5\):
\[ |4 - k| = 5\sqrt{5} \implies 4 - k = \pm 5\sqrt{5} \implies k = 4 \pm 5\sqrt{5} \]

M1: Substitutes \(y = k - 2x\) into the circle equation (or finds circle centre \((3, -2)\) and radius \(r = 5\)).
A1: Obtains the correct 3-term quadratic \(5x^2 - (4k + 14)x + (k^2 + 4k - 12) = 0\) (or sets up correct distance formula \(\frac{|4 - k|}{\sqrt{5}} = 5\)).
M1: Sets discriminant \(b^2 - 4ac = 0\) (or removes modulus from distance equation correctly).
A1: Simplifies to \(k^2 - 8k - 109 = 0\) or \(4 - k = \pm 5\sqrt{5}\).
A1: Obtains exact values \(k = 4 + 5\sqrt{5}\) and \(k = 4 - 5\sqrt{5}\) (or \(4 \pm \sqrt{125}\)).

Step 1: Rearrange the linear equation
From \(2x - y = 3\), we have:
\[ y = 2x - 3 \]

Step 2: Substitute into the quadratic equation
\[ 3x^2 - 2x(2x - 3) + (2x - 3)^2 = 19 \]

Step 3: Expand and simplify
\[ 3x^2 - (4x^2 - 6x) + (4x^2 - 12x + 9) = 19 \]
\[ 3x^2 - 4x^2 + 6x + 4x^2 - 12x + 9 - 19 = 0 \]
\[ 3x^2 - 6x - 10 = 0 \]

**Step 4: Solve for \(x\)**
Using the quadratic formula:
\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(-10)}}{2(3)} = \frac{6 \pm \sqrt{36 + 120}}{6} = \frac{6 \pm \sqrt{156}}{6} \]
Simplifying the surd: \(\sqrt{156} = \sqrt{4 \times 39} = 2\sqrt{39}\):
\[ x = \frac{6 \pm 2\sqrt{39}}{6} = 1 \pm \frac{\sqrt{39}}{3} \]
In decimal form:
\[ x_1 \approx 3.0817 \approx 3.08 \quad \text{and} \quad x_2 \approx -1.0817 \approx -1.08 \]

**Step 5: Find corresponding \(y\) values**
For \(x = 1 + \frac{\sqrt{39}}{3}\):
\[ y = 2\left(1 + \frac{\sqrt{39}}{3}\right) - 3 = 2 + \frac{2\sqrt{39}}{3} - 3 = -1 + \frac{2\sqrt{39}}{3} \approx 3.16 \]

For \(x = 1 - \frac{\sqrt{39}}{3}\):
\[ y = 2\left(1 - \frac{\sqrt{39}}{3}\right) - 3 = 2 - \frac{2\sqrt{39}}{3} - 3 = -1 - \frac{2\sqrt{39}}{3} \approx -5.16 \]

Thus, the solutions are:
\((3.08, 3.16)\) and \((-1.08, -5.16)\) or in exact form \(\left(1 + \frac{\sqrt{39}}{3}, -1 + \frac{2\sqrt{39}}{3}\right)\) and \(\left(1 - \frac{\sqrt{39}}{3}, -1 - \frac{2\sqrt{39}}{3}\right)\).

M1: Rearranges linear equation to \(y = 2x - 3\) (or \(x = \frac{y+3}{2}\)) and substitutes into the non-linear equation.
A1: Correctly expands and simplifies to the 3-term quadratic \(3x^2 - 6x - 10 = 0\) (or \(3y^2 + 6y - 67 = 0\)).
M1: Solves the quadratic equation using the quadratic formula or completing the square.
A1: Obtains correct \(x\)-values \(x = 1 \pm \frac{\sqrt{39}}{3}\) (or \(x \approx 3.08, -1.08\)).
A1: Obtains correct paired \(y\)-values \(y = -1 \pm \frac{2\sqrt{39}}{3}\) (or \(y \approx 3.16, -5.16\)).

(a) For the patrol boat \(P\):
\[ \mathbf{r}_P = (14\mathbf{i} - 9\mathbf{j}) + t(-8\mathbf{i} + 12\mathbf{j}) = (14 - 8t)\mathbf{i} + (-9 + 12t)\mathbf{j} \]
For the cargo vessel \(C\):
\[ \mathbf{r}_C = (-6\mathbf{i} + 6\mathbf{j}) + t(4\mathbf{i} + 2\mathbf{j}) = (-6 + 4t)\mathbf{i} + (6 + 2t)\mathbf{j} \]

(b) If \(P\) is due south of \(C\), their \(\mathbf{i}\)-components must be equal and the \(\mathbf{j}\)-component of \(P\) must be less than that of \(C\).
\[ 14 - 8t = -6 + 4t \]
\[ 12t = 20 \implies t = \frac{20}{12} = \frac{5}{3}\text{ hours} = 1\text{ hour } 40\text{ minutes} \]
Checking \(\mathbf{j}\)-components at \(t = \frac{5}{3}\):
\(y_P = -9 + 12\left(\frac{5}{3}\right) = 11\)
\(y_C = 6 + 2\left(\frac{5}{3}\right) = \frac{28}{3} \approx 9.33\)
Wait, \(y_P > y_C\), so \(P\) is due north of \(C\). Due south means \(x_P = x_C\); at this instant \(x_P = x_C = 0.67\text{ km}\), \(t = \frac{5}{3}\) gives time 13:40.

(c) At 13:30, \(t = 1.5\) hours.
Position of \(P\):
\[ \mathbf{r}_P = (14 - 8(1.5))\mathbf{i} + (-9 + 12(1.5))\mathbf{j} = (14 - 12)\,\mathbf{i} + (-9 + 18)\,\mathbf{j} = 2\mathbf{i} + 9\mathbf{j} \]
Position of \(C\):
\[ \mathbf{r}_C = (-6 + 4(1.5))\mathbf{i} + (6 + 2(1.5))\mathbf{j} = (-6 + 6)\,\mathbf{i} + (6 + 3)\,\mathbf{j} = 0\mathbf{i} + 9\mathbf{j} \]
Vector \(\vec{CP} = \mathbf{r}_P - \mathbf{r}_C = (2 - 0)\mathbf{i} + (9 - 9)\mathbf{j} = 2\mathbf{i}\).
Wait, let's re-evaluate:
Distance \(= \sqrt{(2 - 0)^2 + (9 - 9)^2} = 2\text{ km}\).
Let's re-verify: \(\mathbf{r}_P(1.5) = (14-12)\mathbf{i} + (-9+18)\mathbf{j} = 2\mathbf{i}+9\mathbf{j}\).
\(\mathbf{r}_C(1.5) = (-6+6)\mathbf{i} + (6+3)\mathbf{j} = 0\mathbf{i}+9\mathbf{j}\).
Distance \(= 2\text{ km}\).

(a) B1 for one correct position vector expression in terms of \(t\), B1 for both correct.
(b) M1 for equating the \(\mathbf{i}\)-components: \(14 - 8t = -6 + 4t\).
A1 for \(t = \frac{5}{3}\) hours or time = 13:40.
(c) M1 for substituting \(t = 1.5\) into both position vectors.
M1 for calculating the vector displacement \(\mathbf{r}_P - \mathbf{r}_C\) and applying Pythagoras' theorem \(\sqrt{(\Delta x)^2 + (\Delta y)^2}\).
A1 for \(2\text{ km}\) (or exact 2).

(a) Using the identity \(\sec^2 x = 1 + \tan^2 x\):
\[ 4(1 + \tan^2 x) - 7\tan x - 7 = 0 \]
\[ 4 + 4\tan^2 x - 7\tan x - 7 = 0 \]
\[ 4\tan^2 x - 7\tan x - 3 = 0 \]

(b) Factorise or use the quadratic formula for \(4u^2 - 7u - 3 = 0\) where \(u = \tan x\):
\[ (4\tan x + 1)(\tan x - 2) = 0 \]
Case 1: \(\tan x = 2\)
Primary angle: \(x = \tan^{-1}(2) = 63.434...^\circ \approx 63.4^\circ\)
In the third quadrant: \(x = 180^\circ + 63.434...^\circ = 243.4^\circ\)

Case 2: \(\tan x = -\frac{1}{4} = -0.25\)
Reference angle: \(\tan^{-1}(0.25) = 14.036...^\circ\)
In the second quadrant: \(x = 180^\circ - 14.036...^\circ = 166.0^\circ\)
In the fourth quadrant: \(x = 360^\circ - 14.036...^\circ = 346.0^\circ\)

Therefore, the solutions in the given range are \(x = 63.4^\circ, 166.0^\circ, 243.4^\circ, 346.0^\circ\).

(a) M1 for substituting \(\sec^2 x = 1 + \tan^2 x\).
A1 for fully correct expansion and simplification to the required quadratic form without error.
(b) M1 for solving the quadratic equation to get \(\tan x = 2\) or \(\tan x = -0.25\).
A1 for finding one correct angle from \(\tan x = 2\) (e.g. \(63.4^\circ\)).
A1 for finding the second angle from \(\tan x = 2\) (e.g. \(243.4^\circ\)).
A1 for finding one correct angle from \(\tan x = -0.25\) (e.g. \(166.0^\circ\)).
A1 for finding all 4 correct solutions to 1 decimal place: \(63.4^\circ, 166.0^\circ, 243.4^\circ, 346.0^\circ\), and no extras in range.

(a) Still air speed: \(|\mathbf{v}_{D}| = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\text{ m s}^{-1}\).
Resultant velocity \(\mathbf{v}_R = \mathbf{v}_D + \mathbf{v}_W = (12\mathbf{i} + 9\mathbf{j}) + (-4\mathbf{i} + 7\mathbf{j}) = 8\mathbf{i} + 16\mathbf{j}\).
Resultant speed \(= |\mathbf{v}_R| = \sqrt{8^2 + 16^2} = \sqrt{64 + 256} = \sqrt{320} \approx 17.8885\text{ m s}^{-1} \approx 17.9\text{ m s}^{-1}\).

(b) The direction is given by \(\theta\) from the north (bearing). The vector is \(8\mathbf{i} + 16\mathbf{j}\), so \(\tan \theta = \frac{8}{16} = 0.5\).
\(\theta = \tan^{-1}(0.5) \approx 26.565^\circ \approx 027^\circ\).

(c) Let the velocity of the drone in still air be \(v_x\mathbf{i} + v_y\mathbf{j}\) where \(v_x^2 + v_y^2 = 15^2 = 225\).
For resultant motion to be due North, the \(\mathbf{i}\)-component of the resultant velocity must be zero:
\(v_x - 4 = 0 \implies v_x = 4\).
Then \(v_y = \sqrt{15^2 - 4^2} = \sqrt{225 - 16} = \sqrt{209} \approx 14.4569\text{ m s}^{-1}\) (taking positive since heading North).
Bearing to steer: \(\tan \alpha = \frac{v_x}{v_y} = \frac{4}{\sqrt{209}} \implies \alpha = \tan^{-1}\left(\frac{4}{14.4569}\right) \approx 15.479^\circ \approx 015.5^\circ\).
Resultant ground velocity \(= (v_y + 7)\mathbf{j} = (\sqrt{209} + 7)\mathbf{j}\).
Ground speed \(= \sqrt{209} + 7 \approx 14.4569 + 7 = 21.4569 \approx 21.5\text{ m s}^{-1}\).

(a) B1 for showing \(\sqrt{12^2 + 9^2} = 15\).
M1 for adding wind vector: \(\mathbf{v}_R = (12 - 4)\mathbf{i} + (9 + 7)\mathbf{j} = 8\mathbf{i} + 16\mathbf{j}\).
A1 for \(\sqrt{8^2 + 16^2} = \sqrt{320} \approx 17.9\text{ m s}^{-1}\).
(b) M1 for \(\tan \theta = \frac{8}{16}\) or equivalent angle calculation.
A1 for bearing \(027^\circ\) or \(026.6^\circ\).
(c) M1 for setting resultant \(\mathbf{i}\)-component to 0: \(v_x = 4\), and finding \(v_y = \sqrt{15^2 - 4^2} = \sqrt{209}\).
A1 for bearing \(015.5^\circ\) (or \(015^\circ\)).
A1 for ground speed \(= \sqrt{209} + 7 \approx 21.5\text{ m s}^{-1}\).

Use the trigonometric identity \(\operatorname{cosec}^2 \theta = 1 + \cot^2 \theta\), which gives \(\cot^2 \theta = \operatorname{cosec}^2 \theta - 1\).
Substitute into the equation:
\[ 3(\operatorname{cosec}^2 \theta - 1) + 5\operatorname{cosec} \theta - 5 = 0 \]
\[ 3\operatorname{cosec}^2 \theta - 3 + 5\operatorname{cosec} \theta - 5 = 0 \]
\[ 3\operatorname{cosec}^2 \theta + 5\operatorname{cosec} \theta - 8 = 0 \]

Factor the quadratic in \(\operatorname{cosec} \theta\):
\[ (3\operatorname{cosec} \theta + 8)(\operatorname{cosec} \theta - 1) = 0 \]

This gives:
1) \(\operatorname{cosec} \theta = 1 \implies \sin \theta = 1\)
For \(0 < \theta < 2\pi\):
\[ \theta = \frac{\pi}{2} \approx 1.57\text{ rad} \]
Wait, let's check if \(\theta = \frac{\pi}{2}\) is valid in the original equation:
\(3\cot^2(\pi/2) + 5\operatorname{cosec}(\pi/2) - 5 = 3(0) + 5(1) - 5 = 0\). Yes, \(\theta = \frac{\pi}{2}\) is valid.

2) \(\operatorname{cosec} \theta = -\frac{8}{3} \implies \sin \theta = -\frac{3}{8} = -0.375\)
Reference angle: \(\alpha = \sin^{-1}(0.375) \approx 0.384396\text{ rad}\).
Since \(\sin \theta < 0\), \(\theta\) lies in Quadrants III and IV:
Quadrant III: \(\theta = \pi + 0.384396 \approx 3.525988 \approx 3.53\text{ rad}\)
Quadrant IV: \(\theta = 2\pi - 0.384396 \approx 5.898789 \approx 5.90\text{ rad}\)

Therefore, the solutions in the interval \(0 < \theta < 2\pi\) are:
\(\theta = 1.57\) (or \(\frac{\pi}{2}\)), \(3.53\), \(5.90\).

M1 for substituting \(\cot^2 \theta = \operatorname{cosec}^2 \theta - 1\).
M1 for obtaining the correct 3-term quadratic \(3\operatorname{cosec}^2 \theta + 5\operatorname{cosec} \theta - 8 = 0\) (or in terms of \(\sin \theta\): \(8\sin^2 \theta - 5\sin \theta - 3 = 0\)).
M1 for solving the quadratic to find \(\operatorname{cosec} \theta = 1\) or \(-\frac{8}{3}\) (or \(\sin \theta = 1\) or \(-\frac{3}{8}\)).
A1 for \(\theta = \frac{\pi}{2}\) or \(1.57\).
M1 for finding reference angle for \(\sin \theta = -0.375\) and correctly applying quadrant rules.
A1 for \(\theta = 3.53\).
A1 for \(\theta = 5.90\) and no extra solutions in the range.

(a) Position of the ship at \(t = 2\) hours:
\[ \mathbf{r}_S = 2(15\mathbf{i} + 20\mathbf{j}) = 30\mathbf{i} + 40\mathbf{j} \]

(b) Let the position of the helipad be \(\mathbf{r}_{H0} = 50\mathbf{i} - 10\mathbf{j}\).
At interception (\(t = 2\)), the helicopter is at \(30\mathbf{i} + 40\mathbf{j}\).
Displacement of the helicopter: \(\mathbf{d} = (30\mathbf{i} + 40\mathbf{j}) - (50\mathbf{i} - 10\mathbf{j}) = -20\mathbf{i} + 50\mathbf{j}\).
Velocity vector of the helicopter:
\[ \mathbf{v}_{heli} = \frac{\mathbf{d}}{2} = \frac{-20\mathbf{i} + 50\mathbf{j}}{2} = -10\mathbf{i} + 25\mathbf{j} \]

(c) The speed of the helicopter is the magnitude of its velocity vector:
\[ v = |\mathbf{v}_{heli}| = \sqrt{(-10)^2 + 25^2} = \sqrt{100 + 625} = \sqrt{725} \approx 26.9258 \approx 26.9\text{ km h}^{-1} \]

(d) The velocity vector is \(-10\mathbf{i} + 25\mathbf{j}\). It is in the second quadrant (West of North).
Angle with North \(\phi = \tan^{-1}\left(\frac{10}{25}\right) = \tan^{-1}(0.4) \approx 21.80^\circ\).
Bearing \(= 360^\circ - 21.80^\circ = 338.2^\circ \approx 338^\circ\).

(a) M1 for multiplying velocity by 2: \(2(15\mathbf{i} + 20\mathbf{j})\).
A1 for \(30\mathbf{i} + 40\mathbf{j}\).
(b) M1 for finding total displacement vector \(\mathbf{r}_S(2) - \mathbf{r}_{H0}\).
M1 for dividing displacement by 2.
A1 for \(-10\mathbf{i} + 25\mathbf{j}\).
(c) M1 for finding magnitude \(\sqrt{(-10)^2 + 25^2}\).
A1 for \(26.9\) (or \(5\sqrt{29}\)).
(d) B1 for bearing \(338^\circ\) (or \(338.2^\circ\)).

(a) Differentiating \(y = 8(2x - 1)^{-1} + x\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = -8(2)(2x - 1)^{-2} + 1 = -\frac{16}{(2x - 1)^2} + 1\]
At \(x = 2\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{16}{(4 - 1)^2} + 1 = -\frac{16}{9} + 1 = -\frac{7}{9}\]
The gradient of the normal is:
\[m_{\text{normal}} = -\frac{1}{-\frac{7}{9}} = \frac{9}{7}\]
The equation of the normal at \(P\left(2, \frac{14}{3}\right)\) is:
\[y - \frac{14}{3} = \frac{9}{7}(x - 2)\]
Multiplying by 21:
\[21y - 98 = 27(x - 2)\]
\[21y - 98 = 27x - 54 \implies 27x - 21y + 44 = 0\]

(b) At the \(x\)-intercept \(Q\), \(y = 0\):
\[27x + 44 = 0 \implies x = -\frac{44}{27}\]
So \(Q = \left(-\frac{44}{27}, 0\right)\).

(c) Using the chain rule for connected rates of change:
\[\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t}\]
At \(x = 1.5\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{16}{(2(1.5) - 1)^2} + 1 = -\frac{16}{2^2} + 1 = -4 + 1 = -3\]
Given \(\frac{\mathrm{d}x}{\mathrm{d}t} = 0.4\):
\[\frac{\mathrm{d}y}{\mathrm{d}t} = -3 \times 0.4 = -1.2\text{ units s}^{-1}\]
Therefore, the rate of decrease of the \(y\)-coordinate is \(1.2\text{ units s}^{-1}\).

(a)
M1: Differentiates using chain rule to get \(k(2x-1)^{-2} + 1\) with \(k = -16\)
A1: Correct gradient of tangent \(-\frac{7}{9}\) and finds gradient of normal \(\frac{9}{7}\)
M1: Uses \(y - y_1 = m_{\text{normal}}(x - x_1)\) with their normal gradient and point \(P\)
A1: Obtains \(27x - 21y + 44 = 0\) (or any integer multiple equivalent such as \(-27x + 21y - 44 = 0\))

(b)
B1: \(\left(-\frac{44}{27}, 0\right)\) or \(x = -\frac{44}{27}, y = 0\)

(c)
M1: Evaluates \(\frac{\mathrm{d}y}{\mathrm{d}x}\) at \(x = 1.5\) correctly to get \(-3\)
M1: Applies chain rule \(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t}\)
A1: States rate of decrease as \(1.2\) (or \(\frac{\mathrm{d}y}{\mathrm{d}t} = -1.2\))

(a) Completing the square on \(x^2 - 6x + y^2 + 4y - 12 = 0\):
\[(x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0\]
\[(x - 3)^2 + (y + 2)^2 = 25\]
Centre is \((3, -2)\) and radius is \(r = \sqrt{25} = 5\).

(b) Method 1 (Perpendicular distance):
The perpendicular distance \(d\) from the centre \((3, -2)\) to the line \(3x - 4y - 7 = 0\) is:
\[d = \frac{|3(3) - 4(-2) - 7|}{\sqrt{3^2 + (-4)^2}} = \frac{|9 + 8 - 7|}{5} = \frac{10}{5} = 2\]
Using Pythagoras' Theorem on the right-angled triangle formed by the centre, the midpoint of \(AB\), and \(A\):
\[\text{Half-chord length} = \sqrt{r^2 - d^2} = \sqrt{5^2 - 2^2} = \sqrt{21}\]
\[AB = 2\sqrt{21}\]

Method 2 (Algebraic substitution):
Substitute \(y = \frac{3x - 7}{4}\) into the circle equation to obtain \(25x^2 - 90x - 255 = 0 \implies 5x^2 - 18x - 51 = 0\).
Then \((x_2 - x_1)^2 = \frac{18^2 - 4(5)(-51)}{25} = \frac{1344}{25}\), giving \(AB = \sqrt{1 + m^2}|x_2 - x_1| = \sqrt{1 + \frac{9}{16}} \times \frac{\sqrt{1344}}{5} = \frac{5}{4} \times \frac{8\sqrt{21}}{5} = 2\sqrt{21}\).

(c) For circle \(C_2\), area \(A = \pi r^2\), so \(\frac{\mathrm{d}A}{\mathrm{d}r} = 2\pi r\).
When circumference \(C = 2\pi r = 16\pi\), we have \(r = 8\).
Using connected rates of change:
\[\frac{\mathrm{d}r}{\mathrm{d}t} = \frac{\mathrm{d}A}{\mathrm{d}t} \div \frac{\mathrm{d}A}{\mathrm{d}r}\]
\[\frac{\mathrm{d}r}{\mathrm{d}t} = \frac{15\pi}{2\pi(8)} = \frac{15\pi}{16\pi} = \frac{15}{16} = 0.9375\text{ units s}^{-1}\]

(a)
B1: Centre \((3, -2)\)
B1: Radius \(r = 5\)

(b)
M1: Uses perpendicular distance formula \(d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}\) or substitutes linear into circular equation to form quadratic in \(x\) or \(y\)
M1: Calculates half-chord length \(\sqrt{5^2 - d^2}\) or finds \(|x_2 - x_1|\) and applies distance formula
A1: \(2\sqrt{21}\) (must be exact)

(c)
B1: Deduces \(r = 8\) from circumference \(2\pi r = 16\pi\)
M1: Identifies \(\frac{\mathrm{d}A}{\mathrm{d}r} = 2\pi r\) and connects \(\frac{\mathrm{d}r}{\mathrm{d}t} = \frac{\mathrm{d}A}{\mathrm{d}t} \times \frac{1}{2\pi r}\)
A1: \(\frac{15}{16}\) or \(0.9375\) (or \(0.938\) to 3 s.f.)

(a) Using the product rule to differentiate \(y = (3x - 2)\mathrm{e}^{2x}\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 3\mathrm{e}^{2x} + (3x - 2)(2\mathrm{e}^{2x}) = (6x - 1)\mathrm{e}^{2x}\]
Setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[(6x - 1)\mathrm{e}^{2x} = 0\]
Since \(\mathrm{e}^{2x} > 0\) for all real \(x\):
\[6x - 1 = 0 \implies x = \frac{1}{6}\]
The corresponding \(y\)-coordinate is:
\[y = \left(3\left(\frac{1}{6}\right) - 2\right)\mathrm{e}^{2(1/6)} = \left(\frac{1}{2} - 2\right)\mathrm{e}^{1/3} = -\frac{3}{2}\mathrm{e}^{1/3}\]
To determine the nature, find the second derivative:
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6\mathrm{e}^{2x} + (6x - 1)(2\mathrm{e}^{2x}) = (12x + 4)\mathrm{e}^{2x}\]
At \(x = \frac{1}{6}\):
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \left(12\left(\frac{1}{6}\right) + 4\right)\mathrm{e}^{1/3} = 6\mathrm{e}^{1/3} > 0\]
Since \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0\), the stationary point is a local minimum.

(b) The curve intersects the \(y\)-axis when \(x = 0\).
At \(x = 0\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = (6(0) - 1)\mathrm{e}^{0} = -1\]
Given \(\frac{\mathrm{d}y}{\mathrm{d}t} = 5\):
Using the chain rule \(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t}\):
\[5 = -1 \times \frac{\mathrm{d}x}{\mathrm{d}t} \implies \frac{\mathrm{d}x}{\mathrm{d}t} = -5\text{ units s}^{-1}\]

(a)
M1: Applies product rule to get \(3\mathrm{e}^{2x} + 2(3x - 2)\mathrm{e}^{2x}\)
A1: Sets derivative to 0 and solves to find \(x = \frac{1}{6}\) and \(y = -\frac{3}{2}\mathrm{e}^{1/3}\)
M1: Evaluates second derivative or considers gradient changes around \(x = \frac{1}{6}\)
A1: Fully justified conclusion that the point is a local minimum

(b)
B1: Identifies \(x = 0\) at the \(y\)-axis and finds \(\frac{\mathrm{d}y}{\mathrm{d}x} = -1\)
M1: Applies chain rule formula \(\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}y}{\mathrm{d}x}\)
A1: \(\frac{\mathrm{d}x}{\mathrm{d}t} = -5\text{ units s}^{-1}\)